Assuming the following relationship has been demonstrated

$$r=\frac{m u_{0}\sin \theta_0}{qB\sqrt{1-\left(\dfrac{u_{0}^2}{c^2}\right)}}=\frac{p_0\sin \theta_0}{qB}$$

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where $p_0=mu_0/\sqrt{1-\beta^2}$ represents the relativistic momentum of the particle. In the relativistic case, therefore, $\omega$ (relativistic angular velocity $\omega=\frac{qB}{m} \sqrt{1-\left(\frac{u_{0}^2}{c^2}\right)^2}$) is no longer constant but depends on the speed $\bar{u}_{0}$ of the charged particle $q$; in fact the factor $\gamma$ is present.

The step $p$ (different of the momentum $p_0$) is obtained from the product of the parallel component of the initial particle speed $u_{0\parallel}$ and the period $T$:


Making some considerations about the angle $\theta_0$, if $\theta_0=0$ why I obtain a straight line? Don't you have $r=0$?

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    $\begingroup$ Why do some users want to close my question? Is there a reason? $\endgroup$ – Sebastiano Jul 12 '19 at 22:21
  • $\begingroup$ It isn’t clear what $r$ is the radius of, what $\theta_0$ is the angle between, what $\omega$ is supposed to be, and how $\bar{u}$, $u_0$, and $u$ differ. $\endgroup$ – G. Smith Jul 12 '19 at 22:27
  • $\begingroup$ Also, in relativity, $u_0$ and $p_0$ usually mean the covariant time component of the four-momentum, but I don’t think you’re using them that way. $\endgroup$ – G. Smith Jul 12 '19 at 22:30
  • $\begingroup$ Finally, I don’t think you mean “$\omega$ is no longer constant”. You seem to mean “$\omega$ is no longer speed-independent”. $\endgroup$ – G. Smith Jul 12 '19 at 22:32
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    $\begingroup$ So, overall, the question is quite unclear. $\endgroup$ – G. Smith Jul 12 '19 at 22:35

When $\theta_0$ is zero, the trajectory is no longer a spiral around the $z$-axis; it is a straight line along the $z$-axis, for which $r=0$ in cylindrical coordinates.

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    $\begingroup$ Finally a fair and kind user! It is impossible that in my recent questions I am targeted and how in this question I want to close what I write. I invite you to come to TeX.SE. I signed up to help others but not to have the gift of omnipotence. $\endgroup$ – Sebastiano Jul 12 '19 at 22:55
  • $\begingroup$ Kindest ("Gentilissimo") user thank you once again for your courteous reply. I will be happy to receive your possible answer on my old or recent questions. Thank you for everything and good work. $\endgroup$ – Sebastiano Jul 13 '19 at 10:30

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